# How to Solve the Two-Variable Cubic Equation System?

The difference between cubes of two variables is divided by the difference between variables then we get 19, if the sum of cubes divides with the sum of variables then we get 7, then solve the two-variable cubic equation system

## Algebra math problem: Find the solution to the cubic equation system

\dfrac{x^3-y^3}{x-y}=19\\

\dfrac{x^3+y^3}{x+y}=7

x=?? \ \ \ \ \&\ \ \ \ y=??

## Solution

Let

\begin{aligned} & \dfrac{x^3-y^3}{x-y}=19..............................eq(1)\\ \\ & \dfrac{x^3+y^3}{x+y}=7................................eq(2)\\ \end{aligned}

Factorise *x*³ *−* *y*³ and *x*³ + *y*³ then

*x*³ *−* *y*³ = (*x − y*)(*x*² +* xy *+* y*²)

*x*³ + *y*³ = (*x *+* y*)(*x*² *−* *xy *+* y*²)

From equation 1

\begin{aligned} &\dfrac{x^3-y^3}{x-y}=19\\ \\ &\Rightarrow \ \dfrac{(x-y)(x^2+xy+y^2)}{x-y}=19\\ \\ &\Rightarrow \ x^2+xy+y^2=19, \ \ x-y \neq0\\ \end{aligned}

From equation 2

\begin{aligned} &\dfrac{x^3+y^3}{x+y}=7\\ \\ &\Rightarrow \ \dfrac{(x+y)(x^2-xy+y^2)}{x+y}=7\\ \\ &\Rightarrow \ x^2-xy+y^2=7, \ \ x+y \neq0\\ \end{aligned}

Let

*x*²* *+* xy *+* y*² = 19………..*eq*(3)

*x*²* −* *xy *+* y*² = 7………….*eq*(4)

Add equation 3 and equation 4, then

*x*²* *+* xy *+* y*² + *x*²* −* *xy * +* y*² = 19 + 7

⇒ 2*x*²* *+ 2*y*² = 26

⇒ *x*²* *+ *y*² = 13………….*eq*(5)

Equation 3 minus equation 4, then

*x*²* *+* xy *+* y*² *−* *x*²* *+ *xy −* *y*² = 19 *−* 7

⇒ 2*xy* = 12………….*eq*(6)

Add equation 5 and equation 6

*x*²* *+ *y*² + 2*xy* = 13 + 12

⇒ (*x* + *y*)² = 25

⇒ *x* + *y* = ±5

⇒ *y* = ±5 *−* *x*

When *y* = 5 *−* *x*

From equation 6

2*xy* = 2*x*(5 *−* *x*) = 12

⇒ 5x *−* *x* ²= 6

⇒ *x*² *−* 5*x* + 6= 0

*x*² *−* 5*x* *+* 6 = 0 is a quadratic equation so

\begin{aligned} x&=\dfrac{5\pm\sqrt{(-5)^2-4\times1\times6}}{2\times1}\\ \\ \Rightarrow x&=\dfrac{5\pm\sqrt{25-24}}{2}\\ \\ \Rightarrow x&=\dfrac{5\pm1}{2}\\ \\ \Rightarrow x&=3 \ \ \& \ \ 2\\ \\ \end{aligned}

*y* = 5 *−* *x* so

* y* = 5

*−*3 =

**2**

or, * y* = 5

*−*2 =

**3**

⇒ (*x*, *y*) = (3, 2) & (2, 3)

When *y* = *−*5 *−* *x*

From equation 6

2*xy* = 2*x*(*−*5 *−* *x*) = 12

⇒ *−*5x *−* *x* ²= 6

⇒ *x*² + 5x + 6= 0

### *x*² *+* 5*x* *+* 6 = 0 is a quadratic equation so

\begin{aligned} x&=\dfrac{-5\pm\sqrt{5^2-4\times1\times6}}{2\times1}\\ \\ \Rightarrow x&=\dfrac{-5\pm\sqrt{25-24}}{2}\\ \\ \Rightarrow x&=\dfrac{-5\pm1}{2}\\ \\ \Rightarrow x&=-2 \ \ \& \ -3\\ \\ \end{aligned}

*y* = *−*5 *−* *x* so

* y* =

*−*5

*−*(

*−*2) =

*−*3or, * y* =

*−*5

*−*(

*−*3) =

*−*2⇒ (*x*, *y*) = (** −**2,

**3) & (**

*−***3,**

*−**−*2)

## So solution to the cubic equation

**( x, y) = (3, 2), (2, 3) (−2, −3) & (−3, −2)**